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To solve the equation \(\sqrt{3x - 3} = \sqrt{2x + 8}\) for \(x\), let's follow these steps:

### Step 1: Square Both Sides
Since both sides are square roots, we can eliminate the square roots by squaring both sides of the equation:
\[
(\sqrt{3x - 3})^2 = (\sqrt{2x + 8})^2
\]
This simplifies to:
\[
3x - 3 = 2x + 8
\]

### Step 2: Solve for \(x\)
Now, isolate \(x\) by moving terms involving \(x\) to one side and constants to the other:
\[
3x - 2x = 8 + 3
\]
\[
x = 11
\]

### Step 3: Verify the Solution
To ensure that \(x = 11\) is a valid solution, we substitute it back into the original equation:
\[
\sqrt{3(11) - 3} = \sqrt{2(11) + 8}
\]
Calculating inside the square roots:
\[
\sqrt{33 - 3} = \sqrt{22 + 8}
\]
\[
\sqrt{30} = \sqrt{30}
\]
Since both sides are equal, \(x = 11\) is indeed a solution.

### Conclusion
The correct answer is:
\[
x = 11
\]

Would you like a detailed explanation of each step, or do you have any questions about the solution?

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Here are 5 related questions for further practice:
1. How would you solve an equation with different square root terms, like \(\sqrt{5x + 2} = \sqrt{3x + 14}\)?
2. What happens if you square both sides of an equation but one side has a negative expression?
3. How do you solve an equation with a square root on one side and a linear term on the other, like \(\sqrt{4x - 1} = x + 2\)?
4. Can you find the solution to \(\sqrt{x + 5} = 3\)?
5. What are some potential pitfalls when squaring both sides of an equation?

**Tip:** Always verify solutions when squaring both sides, as this process can sometimes introduce extraneous solutions.

Solve \(\sqrt{3x - 3} = \sqrt{2x + 8}\) for \(x\).

Learn how to solve the radical equation \(\sqrt{3x - 3} = \sqrt{2x + 8}\) with a detailed step-by-step solution. This problem covers algebraic concepts related to square roots and radical equations, suitable for students in Grades 9-11.

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